AQA · OCR A · acid–base and redox
Back titration questions, worked examples and generator
Start with a guided calculation or go straight to a full exam question. Review the method only if you need it.
Step by step
Guided back-titration practice
Work through one amount at a time. I’d label each value as added, left or reacted; that makes the subtraction much harder to reverse.
Give your answer to three significant figures.
Back titration questions and answers
Questions may ask for purity, tablet content, concentration or the identity of a compound. Some use excess NaOH or several redox equations, so read the method before choosing a route.
Printable practice
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What is a back titration?
Find what was left, then subtract it.
amount that reacted=amount initially added−amount left over
A measured excess of reagent reacts with the sample. The later titration tells you how much of that excess remains. Subtracting gives the amount that reacted with the substance being analysed.
Track each amount through the method: what was added, what remained, and therefore what reacted. A modern question may present those stages in an unfamiliar order and may involve more than two reactions.
Back titration calculation diagram
How the calculation works
The values below change whenever you generate a new guided question. The diagram shows the usual excess-reagent idea: some reacts with the sample and the rest is measured later.
- Calculate the amount of reagent originally added in excess.
- Use the later titration to calculate how much reagent was left.
- If the titration used a measured portion of a larger solution, scale from the portion to the whole solution. Older textbooks may call that portion an aliquot.
- Subtract, then use the equation or equations given in the question.
amount added − amount left = amount that reacted
Worked examples
Three versions of the same underlying calculation
Start with the direct acid–base example, then add dilution, then try a redox route. Each example keeps the amount left over separate from the amount that reacted.
Example 1Simple excess acid, no dilution
Question. A 0.500 g limestone sample is treated with 50.0 cm³ of 0.200 mol dm⁻³ HCl. The remaining acid requires 20.0 cm³ of 0.100 mol dm⁻³ NaOH. Calculate the percentage by mass of CaCO₃.
HCl initially added = 0.200 × 50.0 ÷ 1000 = 0.0100 mol
HCl left over = NaOH used = 0.100 × 20.0 ÷ 1000 = 0.00200 mol
HCl that reacted = 0.0100 − 0.00200 = 0.00800 mol
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O, so CaCO₃ = 0.00800 ÷ 2 = 0.00400 mol
Mass CaCO₃ = 0.00400 × 100.1 = 0.400 g; percentage = 0.400 ÷ 0.500 × 100 = 80.0%
Common wrong route: using 0.00200 mol as the amount that reacted. It is the amount left over.
Example 2Volumetric flask and measured portion
Question. A 1.00 g sample containing MgO reacts with 50.0 cm³ of 1.00 mol dm⁻³ HCl. The mixture is made up to 250 cm³. A 25.0 cm³ portion requires 12.80 cm³ of 0.100 mol dm⁻³ NaOH. Calculate the percentage by mass of MgO.
HCl initially added = 1.00 × 50.0 ÷ 1000 = 0.0500 mol
HCl in the 25.0 cm³ portion = 0.100 × 12.80 ÷ 1000 = 0.001280 mol
HCl in the full 250 cm³ = 0.001280 × 10 = 0.01280 mol
HCl that reacted = 0.0500 − 0.01280 = 0.03720 mol
MgO + 2HCl → MgCl₂ + H₂O, so MgO = 0.03720 ÷ 2 = 0.01860 mol
Mass MgO = 0.01860 × 40.3 = 0.7496 g; percentage = 75.0%
Common wrong route: forgetting that the titre describes only one tenth of the solution.
Example 3Redox back titration
Question. A 0.630 g sample containing Na₂SO₃ reacts with 50.0 cm³ of 0.100 mol dm⁻³ iodine. The iodine left over requires 20.0 cm³ of 0.100 mol dm⁻³ thiosulfate. Calculate the percentage by mass of Na₂SO₃.
I₂ initially added = 0.100 × 50.0 ÷ 1000 = 0.00500 mol
S₂O₃²⁻ used = 0.100 × 20.0 ÷ 1000 = 0.00200 mol
I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻, so I₂ left = 0.00200 ÷ 2 = 0.00100 mol
I₂ that reacted with sulfite = 0.00500 − 0.00100 = 0.00400 mol
SO₃²⁻ and I₂ react 1 : 1, so Na₂SO₃ = 0.00400 mol = 0.504 g
Percentage = 0.504 ÷ 0.630 × 100 = 80.0%
Common wrong route: missing the 2 : 1 thiosulfate-to-iodine ratio before doing the subtraction.
When an answer goes wrong
Find the step that went wrong
My answer is above 100%
Check the subtraction direction, the scale factor from the measured portion to the whole solution, and the mole ratio. A value above 100% usually means the amount of analyte has been made too large.
I used the titre amount directly
The titre usually measures the excess reagent left over. Subtract that amount from the reagent initially added to find what reacted with the original sample.
Which equation comes first?
Begin with the reaction involving the titrant in the burette. That calculation gives the amount left over. Then return to the reaction with the original sample.
What scale factor do I use?
Compare the measured portion with the total volumetric-flask volume. A 25.0 cm³ portion from 250 cm³ represents one tenth, so multiply the amount in the portion by 10.
The titre table has several results
Ignore the rough titre. Use concordant accurate titres, normally those within 0.10 cm³, and calculate their mean.
My unit is wrong
Convert cm³ to dm³ before using n = cV. At the end, check whether the question asks for moles, mass, concentration or percentage.
More titration practice
Choose the titration skill you actually need
Why use a back titration?
A direct titration may be unsuitable when a solid is insoluble, reacts slowly near completion or gives a poor endpoint. A measured excess of reagent is allowed to react completely, then the unused excess is determined accurately by a second titration.
The second titration is only meaningful after the sample has finished reacting with the measured excess.
Its colour change marks the endpoint of the second titration, which measures the excess reagent remaining after the first reaction.
Acid–base and redox back titrations
Acid–base and redox back titrations use the same amount accounting. Identify the reagent added in excess, work out what remained, subtract, then use the equation or equations supplied. Redox questions may require two or three reaction ratios before reaching the original sample.